题解 | #平均活跃天数和月活人数#

select
extract(year_month from submit_time) as month
,round(count(distinct uid,date_format(submit_time,'%y%m%d'))/count(distinct uid),2) as avg_active_days #一个用户同一天提交多次
,count(distinct uid) as mau
from exam_record
where submit_time is not null and year(start_time)=2021
group by month

坑在与存在一个用户同一天提交多次的情况 直接distinct 提交日期，没有真正的去重